Question 701: Thermodynamics & KTG

Question

A metal sphere cools at a rate of $1.5^{\circ} \mathrm{C} / \mathrm{min}$ when its temperature is $80^{\circ} \mathrm{C}$. When the temperature of the sphere is $40^{\circ} \mathrm{C}$, its rate of cooling is $0.3^{\circ} \mathrm{C} / \mathrm{min}$. The temperature of the surrounding $\left(	heta_0ight)$ is

Accepted Answer

According to Newton's law of cooling,
$$ \left|\frac{\mathrm{dT}}{\mathrm{dt}}ight|=\mathrm{k}\left(\mathrm{~T}-	heta_0ight) $$
At $\mathrm{T}=80^{\circ}, 1.5=\mathrm{k}\left[80-	heta_0ight]$
At $\mathrm{T}=40^{\circ}, 0.3=\mathrm{k}\left[40-	heta_0ight]$
Dividing (i) and (ii)
$$ \begin{aligned} & \frac{1.5}{0.3}=\frac{80-	heta_0}{40-	heta_0} \ \Rightarrow & 200-5 	heta_0=80-	heta_0 \ \Rightarrow & 4 	heta_0=120 \ \Rightarrow & 	heta_0=30^{\circ} \mathrm{C} \end{aligned} $$

Answer

$30^{\circ} \mathrm{C}$

Answer

$35^{\circ} \mathrm{C}$

Answer

$25^{\circ} \mathrm{C}$

Answer

$27^{\circ} \mathrm{C}$

Thermodynamics & KTG

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